Probability of Dependent Events | MathHelp.com - Free Educational videos for Students in K-12 | Lumos Learning

## Probability of Dependent Events | MathHelp.com - Free Educational videos for Students in k-12

#### Probability of Dependent Events | MathHelp.com - By MathHelp.com

Transcript
00:00 Andrew has eight black socks and four brown socks in
00:04 his sock drawer and it's too dark to tell the
00:06 colours apart . If he chooses one sock and puts
00:10 it on , then chooses another . Sock . Find
00:13 the probability of the following event . Pf brown than
00:16 brown . In other words , were asked to find
00:20 the probability that Andrew chooses a brown sock , then
00:23 another brown sock . It's important to understand that when
00:27 Andrew chooses the first suck , he puts it on
00:30 , so when he chooses the second sock , there's
00:33 one fewer sock in the drawer . Therefore the outcome
00:37 of the first event affects the outcome of the second
00:40 event , which means that these are dependent events .
00:43 To find the probability of dependent events . We first
00:47 find the probability of each event , then we multiply
00:50 the probabilities together just like we did when finding the
00:53 probability of independent events . However , we have to
00:57 be very careful when finding the probabilities . So let's
01:01 first find the probability that Andrew chooses a brown sock
01:05 . What ? Since there are four brown socks in
01:14 the drawer and there are eight plus four or 12
01:18 total socks in the drawer . The probability that Andrew
01:21 chooses a brown sock is four 12 . Which reduces
01:27 to 1/3 . Remember to always reduce if possible .
01:32 When finding probability . Next let's find the probability that
01:36 Andrew chooses a second . Brown sock . Yeah .
01:43 Yeah . Well yes and here's where it gets tricky
01:49 , remember that when Andrew chooses the first sock he
01:52 puts it on . So if he has already successfully
01:55 chosen a brown sock , there are no longer four
01:58 brown socks in the drawer . There are three and
02:02 there are now eight plus three Or 11 total socks
02:08 in the drawer . So the probability that Andrew chooses
02:11 a second brown sock is three 11th . Now to
02:18 find the probability that he chooses a brown sock puts
02:21 it on , then chooses another Brown sock . We
02:24 multiply the probability that Andrew chooses a brown sock one
02:28 third . Mhm . Times the probability that he chooses
02:35 a second . Brown Sock 3 11th notice that we
02:43 can cross cancel the three and 321 and one now
02:48 multiplying across the numerator one times one is one ,
02:52 and multiplying across the denominators , one times 11 is
02:57 11 , so one third times 3/11 is 1/11 which
03:02 means the probability that Andrew chooses a brown sock ,
03:06 puts it on , then chooses another . Brown sock
03:09 is 1/11 .
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